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JEE Main 20209 JanuaryMorning Shift

Exam: JEE Main
Year: 2020
Date: 9 January
Shift: Morning Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2020 · 9 January · Morning Shift · Chemistry · Chemical Kinetics

For the following reactions A 700K → Product A 500K → _ catalyst...

For the following reactions

\text{A}\overset{700\text{K}}{\to}\text{Product}

\text{A}\overset{500\text{K}}{\mathop{\to}_{\text{catalyst}}}\text{Product}

It was found that the

{E}_{a}

is decreased by

30KJ / mol

in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre-exponential factor is same)

Options

A $75KJ / mol$
B $105KJ / mol$
C $135KJ / mol$
D $198KJ / mol$

Correct answer

A. $75KJ / mol$

Step-by-step solution

{K}_{cat} = K

{Ae}^{- \dfrac{{Ea}_{1}}{{RT}_{1}}} = {Ae}^{- \dfrac{{Ea}_{2}}{{RT}_{2}}}

\dfrac{{Ea}_{1}}{{T}_{1}} = \dfrac{{Ea}_{2}}{{T}_{2}}; {Ea}_{1} = {Ea}_{2} - 30

\dfrac{{Ea}_{2} - 30}{500} = \dfrac{{Ea}_{2}}{700}

5{Ea}_{2} = 7{Ea}_{2} - 210

{Ea}_{2} = \dfrac{210}{2} = 105kJ / mole

Activation energy of the catalysed reaction

= 105 - 30 = 75kJ / mole

.

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